This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

2

If $A$ is a Noetherian commutative domain, is the kernel of a map $A^n\to A^m$ always projective?
–
Mariano Suárez-Alvarez♦May 3 '11 at 5:06

Think about k[[x,y,z]] or look up projective dimension of a ring depending upon your tastes. This is true for a ring of projective dimension 1 or 2 though so, in scheme talk, smooth curves or surfaces over a field.
–
Daniel PomerleanoMay 3 '11 at 5:43

Dear Gunnar, no, this doesn't work. The morphism of sheaves $m:\mathcal O \to \mathcal O$ given by multiplication by $x$ is injective, hence the kernel of $m$ is zero and thus free. However if you take the corresponding vector bundle, your $L$, the induced morphism on fibers $m[0]:L[0]\to L[0]$ is no longer injective, as you correctly note. The point is that taking fibers amounts to tensoring with $\mathcal O_0/\frak m_0=\mathbb C$, which is not a flat $\mathcal O_0 $-module and so does not preserve injectivity.
–
Georges ElencwajgMay 3 '11 at 7:17

2

Why was this closed? It is a valid mathematical question, and I was interested in the answer.
–
DrewSep 16 '13 at 17:15

1 Answer
1

Let $X$ be a smooth quasi-projective variety over a field of dimension $n$ and $Z\subseteq X$ a subvariety such that the depth of $\mathscr O_Z$ at a fixed (closed) point of $X$ is $d$. Assume that $n-d\geq 3$, i.e., the projective dimension of $\mathscr O_Z$ at that closed point of $X$ is at least $3$. Then consider a locally free sheaf that surjects onto the ideal sheaf of $Z$. That gives a morphism to $\mathscr O_X$ and its kernel cannot be locally free, because then the projective dimension of $\mathscr O_Z$ at that closed point of $X$ would be $2$.

On the other hand, if your morphism is surjective, then what you want is true. Just count the rank of stalks.